Suppose
$$S_1=1,S_{n+1}=S_n+\frac1{S_n}-\sqrt 2$$
Prove that $S_n$ converges.
I was hinted to observe $S_{2k+1}$ and $S_{2k+2}$ respectively, so I tried calculating $$S_{n+2}-S_n=\frac{(1-\sqrt 2 S_n)^3}{S_n(S_n^2-\sqrt 2 S_n+1)}$$ This is a bit frustrating, since no immediate indication about the sign can be found in the result.
Maybe after sufficient computation (which I haven't given up on yet) things will begin to be clear, but before that, may I ask whether there is any alternative (elegant ones preferred) to this problem? Thanks in advance.
Lemma: if $1>x>\dfrac{1}{\sqrt{2}}$,then $$f(f(x))>\dfrac{1}{\sqrt{2}}$$where $f(x)=x+\dfrac{1}{x}-\sqrt{2}$.
Proof: since $f(x)=x+\dfrac{1}{x}-\sqrt{2}$ is Monotone decreasing function.so $$f(f(x))>f\left(f\left(\dfrac{1}{\sqrt{2}}\right)\right)=\dfrac{1}{\sqrt{2}}$$
since $$S_{n+2}=S_{n+1}+\dfrac{1}{S_{n+1}}-\sqrt{2}=S_{n}+\dfrac{1}{S_{n}}-\sqrt{2} +\dfrac{1}{S_{n}+\dfrac{1}{S_{n}}-\sqrt{2}}-\sqrt{2}$$ first we prove $S_{1}>S_{3}>\cdots>S_{2n+1}$. Let $f(x)=x+\dfrac{1}{x}-\sqrt{2}$,then we have $$S_{2n+1}=f(f(S_{2n-1}))$$ $$S_{2n+1}-S_{2n-1}=\dfrac{-2\sqrt{2}S^3_{2n-1}+6S^2_{2n-1}-3\sqrt{2}S_{2n-1}+1}{S_{2n-1}(S^2_{2n-1}-\sqrt{2}S_{2n-1}+1)}=\dfrac{(1-\sqrt{2}S_{2n-1})^3}{S_{2n-1}(S^2_{2n-1}-\sqrt{2}S_{2n-1}+1)}$$ use Lemma. we can use induction prove $$S_{2n+1}>\dfrac{1}{\sqrt{2}},n\ge 1$$ so $$\dfrac{1}{\sqrt{2}}<\cdots<S_{2n+1}<S_{2n-1}<\cdots<S_{3}<S_{1}$$ so $$\{S_{2n+1}\}$$converges.and we easy to $$\lim_{k\to\infty}S_{2k+1}=\dfrac{1}{\sqrt{2}}$$ the same as $S_{2}<S_{4}<\cdots<S_{2n}<\cdots<\dfrac{1}{\sqrt{2}}$ so $$\{S_{2n}\}$$ converges.and $$\lim_{k\to+\infty}S_{2k}=\dfrac{1}{\sqrt{2}}$$
Add some calculation
\begin{align*}S_{2n+1}-S_{2n-1}&=\dfrac{1}{S_{2n-1}}-2\sqrt{2}+\dfrac{1}{s_{2n-1}+\dfrac{1}{S_{2n-1}}-\sqrt{2}}\\ &=\dfrac{1-2\sqrt{2}S_{2n-1}}{S_{2n-1}}+\dfrac{S_{2n-1}}{S^2_{2n-1}-\sqrt{2}S_{2n-1}+1}\\ &=\dfrac{(1-2\sqrt{2}S_{2n-1})(S^2_{2n-1}-\sqrt{2}S_{2n-1}+1)+S^2_{2n-1}}{S_{2n-1}(S^2_{2n-1}-\sqrt{2}S_{2n-1}+1)}\\ &=\dfrac{S^2_{2n-1}-\sqrt{2}S_{2n-1}+1-2\sqrt{2}S^3_{2n-1}+4S^2_{2n-1}-2\sqrt{2}S_{2n-1}+S^2_{2n-1}}{S_{2n-1}(S^2_{2n-1}-\sqrt{2}S_{2n-1}+1)}\\ &=\dfrac{-2\sqrt{2}S^3_{2n-1}+6S^2_{2n-1}-3\sqrt{2}S_{2n-1}+1}{S_{2n-1}(S^2_{2n-1}-\sqrt{2}S_{2n-1}+1)}\\ &=\dfrac{(1-\sqrt{2}S_{2n-1})^3}{S_{2n-1}(S^2_{2n-1}-\sqrt{2}S_{2n-1}+1)} \end{align*}